rannsan
Find the solution to this system using the substitution method. 2x 5y = 1 and x 3y = 3



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mathslover
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2x5y = 1
2x = 1 + 5y
x = \(\large{\frac{1+5y}{2}}\)

mathslover
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put this value in the second eqn :
\[\large{x3y=3}\]
\[\large{\frac{1+5y}{2}3y=3}\]

matricked
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form the second equation we have x=3y 3 putting it in the first
we have 2(3y3)5y=1
or 6y 65y =1 or y=61
or y=5
hence x=3(5) 3=12
x=12 and y=5

mathslover
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\[\large{\frac{1+5y6y}{2}=3}\]
\[\large{1y=6}\]
\[\large{1+6=y}\]
\[\large{5=y}\]

chemENGINEER
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its too sticky and icky...mmmmmm

mathslover
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similarly find x
x 3(5) = 3
x15=3
x = 3+15

mathslover
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got it @rannsan ?

rannsan
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I think so

rannsan
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Thank you!!